Решение

a)

$$\begin{aligned} \left(\frac{x}{y^2} - \frac{1}{x}\right) : \left(\frac{1}{y} + \frac{1}{x}\right) &= \frac{x^2 - y^2}{xy^2} : \frac{x+y}{xy} = \frac{(x-y)(x+y)}{xy^2} \cdot \frac{xy}{x+y} = \frac{(x-y)(x+y)xy}{xy^2(x+y)} = \frac{x-y}{y} \end{aligned}$$

Ответ: $$\frac{x-y}{y}$$

б)

$$\begin{aligned} \left(\frac{a}{m^2} + \frac{a^2}{m^3}\right) : \left(\frac{m^2}{a^2} + \frac{m}{a}\right) &= \frac{am + a^2}{m^3} : \frac{m^2+ma}{a^2} = \frac{a(m+a)}{m^3} \cdot \frac{a^2}{m(m+a)} = \frac{a^3(m+a)}{m^4(m+a)} = \frac{a^3}{m^4} \end{aligned}$$

Ответ: $$\frac{a^3}{m^4}$$

в)

$$\begin{aligned} \frac{ab+b^2}{3} : \frac{b^3}{3a} + \frac{a+b}{b} &= \frac{b(a+b)}{3} \cdot \frac{3a}{b^3} + \frac{a+b}{b} = \frac{3ab(a+b)}{3b^3} + \frac{a+b}{b} = \frac{a(a+b)}{b^2} + \frac{a+b}{b} = \frac{a(a+b) + b(a+b)}{b^2} = \frac{a^2+ab+ab+b^2}{b^2} = \frac{a^2+2ab+b^2}{b^2} = \frac{(a+b)^2}{b^2} \end{aligned}$$

Ответ: $$\frac{(a+b)^2}{b^2}$$

г)

$$\begin{aligned} \frac{x-y}{x} - \frac{5y}{x^2} \cdot \frac{x^2-xy}{5y} &= \frac{x-y}{x} - \frac{5y}{x^2} \cdot \frac{x(x-y)}{5y} = \frac{x-y}{x} - \frac{5yx(x-y)}{5yx^2} = \frac{x-y}{x} - \frac{x-y}{x} = 0 \end{aligned}$$

Ответ: $$0$$